New answers tagged

0

This is not an answer, and don't take it as such. From a theoretical point of view, you have $A=\sum_{k=1}^{n}\lambda_k P_k$ where the $P_k$ are orthogonal projections. If you start iterating in $A$, or perform various functions of $A$, the problem is that $f(A)=\sum_{k=1}^{n}f(\lambda_k)P_k$. You can get down below the level of a projection. You can form ...


0

Idea for an indirect proof: Suposse unit ball is compact. Cover it by $\cup_{x\in B_1 (0)} B_{\frac{1}{2}} (x) $. By compactness, there exist points finitely many points $a_i, i=1,\dots,n$ such that balls of radius $\frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.


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If it is infinite dimensional, we can find an infinite orthonormal basis, $(e_n)_{n \in \Bbb{N}}$. Notice $||e_j|| = 1$ and $||e_i - e_j|| = \sqrt{2}$ for $i \neq j$. (This is a very standard construction.)


1

Yes, but you have to verify this. Suppose $x=\sum f(v)v\in\operatorname{span}\beta$, where $v\in\beta$ and the sum is finite. Show that $\Vert Tx\Vert^2=\Vert x\Vert^2= \sum|f(v)|^2$ (use orthonormality of $\beta$). Thus, if $Tx=0$, we must have $x=0$, which means that $T$ is injective.


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Note: There is nothing about completeness of $\mathcal{H}$ needed to carry out of the following steps. Because $P_n$ is monotone, then $(P_nx,x)$ is monotone in $n$ for each fixed $x$, and is bounded above by $(x,x)$, which forces convergence of $\lim_n(P_n x,x)$ for all $x$. Then, using polarization, the following expression must also have a limit in $n$ ...


2

Example, $$ A = \frac{1}{i}\frac{d}{dx} $$ on the domain $\mathcal{D}(A)$ of absolutely continuous functions $f \in L^2[0,1]$ for which $f' \in L^2[0,1]$ and $f(0)=0$. Then $A^*$ is the same as $A$ except that the condition $f(0)=0$ is replaced by $f(1)=0$. Then $A^{\star\star}=A$ because $A$ is closed and densely-defined. However, ...


0

Consider the two cases of $V_n$ being monotone increasing and monotone decreasing in $n$. In the first case you have $P_n P_{n+1}=P_n$ and in the second you have $P_n P_{n+1}=P_{n+1}$. At any rate you have either $P_n(P_{n+1}- P_{n})=0$ or $P_{n+1}(P_{n+1}-P_{n})=0$. First consider $V_n$ is increasing. So $(P_{n+1}-P_n)(z)$ is in $V_n^\perp$. This means ...


1

Assume $V_n \subseteq V_{n+1}$ for all $n$. Then $P_{n+1}P_n=P_n$. Using adjoint, $P_nP_{n+1}=P_n$ must also hold because $P_k^*=P_k$ for all $k$. Therefore, for all $x$, $$ (P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x). $$


1

In any infinite-dimensional Hilbert space $\mathcal H$, an orthonormal basis is not a basis in the algebraic sense. Suppose not: let $B$ be such an orthonormal basis. Let $b_1, b_2, \ldots$ be a sequence of distinct members of $B$ and $x = \sum_{j=1}^\infty b_j/j$. If $x = \sum_{b \in B} c_b b$ (where only finitely many $c_b \ne 0$), then $c_b = \langle b, ...


1

Yes, it is fine if you interpret $\langle D^2 f(x) , e_n \rangle$ as the bilinear form $$(y,z) \mapsto \langle D^2 f(x)[y,z], e_n \rangle.$$ As you already said, this follows simply from the chain rule and the linearity of $L_n$.


2

If $s \in S$, then $\mathcal{N}_s = \{ x \in H : \langle x,s \rangle = 0 \}$ is the null space of a continuous linear functional, which is the inverse image of $\{0\}$ under this continuous linear functional. Hence $\mathcal{N}_s$ is closed, as is the intersection $$ S^{\perp} = \bigcap_{s\in S}\mathcal{N}_s. $$


1

Let $T$ be the Fréchet derivative of $f$ at $e\in E$. Then $$ \frac{|f_n(e+h)-f_n(e)-\langle Th,e_n\rangle|}{\|h\|} =\frac{|\langle f(e+h),e_n\rangle-\langle f(e),e_n\rangle-\langle Th,e_n\rangle}{\|h\|} =\frac{|\langle f(e+h)- f(e)- Th,e_n\rangle}{\|h\|} \leq\frac{\| f(e+h)- f(e)- Th\|}{\|h\|}\to0. $$ So the derivative of $f_n$ is $h\longmapsto \langle ...


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Yes, this is true. Indeed, let $t_n$ be a sequence in $S^{\perp}$ and let $t$ be its limit. We want to show that $t \in S^{\perp}$. By the continuity of the inner product, we have that, for all $s \in S$ $$ \langle t, s \rangle = \lim_{n \to \infty} \langle t_n, s \rangle=0 $$ so that $t \in S^{\perp}$. The equality $(S^{\perp})^{\perp}=S$ only holds if ...


1

It's true that $S^\perp$ is closed. This is trivial from the definition. This does not imply that $S$ is closed, because in general $S\ne S^{\perp\perp}$. In fact $S=S^{\perp\perp}$ if and only if $S$ is a closed subspace of $H$; in general $S^{\perp\perp}$ is the closure of the span of $S$.


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It's basically the same, it's mainly a matter of implicit identifications. Namely, in a Hilbert space $H$ there is a natural isomorphism $H\to H^*$ (which is the bra-ket duality), so $H\otimes H\simeq H\otimes H^*$. Now there is a map $H\otimes H^*\to L(H)$ which corresponds to what you call the outer product. So for two kets $|\psi\rangle$ and ...


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The first term contains a typo, it should read $B(x,1/2)$ instead of $B(x + 1/2)$. Moreover, $F + B(0,1/2)$ is the Minkowski sum, which is defined via $$F + B(0,1/2) = \{x + y \colon x \in F, y \in B(0,1/2)\}.$$ Hence, this equality is essentially the definition of the Minkowski sum.


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No it's not. For instance, on $\mathbb{C}^n$, take any matrix $M\in M_n(\mathbb{C})$ that is not hermitian, and put $(X,Y)\mapsto X^t M \overline{Y}$.


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This seems false. Take $U=H$, $Z=Id$, and identify $L(H,\mathbb{R})$ with $H$. Take and orthogonal base $e_1,e_2\ldots,$ and define $f(e_1)=e_2$, $f(e_i)=0$ for $i>1$, extend by linearity. Then $$ \langle e_2,f(e_1)\rangle=1\neq0\langle e_1,f(e_2)\rangle. $$


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It seems that the uniqueness part has already been settled. In your argument regarding the existence part, you have proved that if the sequence $(x_n)_{n=1}^\infty$ you have constructed converges, then its limit belongs to $C$. But so far you have only that $\|x_n\| \rightarrow s$. It remains to be shown that $(x_n)_{n=1}^\infty$ does converge. To this end, ...


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There are two cases: $0\in C$ and $0\notin C$. In the first case the minimum norm is $=0$, and $x_0=0$ is the unique element with this property. If $x_0$ with minimum norm has already been found (and your question indicates that this is already known to you) assume $x_1\neq x_0$ has the same norm and it is an element of $C$. It is easy to see that $x_1$ and ...


1

It's true that the $\ell^p$ and $\ell^q$, $p\ne q$, are not Lipschitz homeomorphic. (And changing a norm to an equivalent one does not change this.) This fact is not something that can be proved with bare hands, unless one of $p,q$ is $\infty$, when separability makes the distinction clear. Chapter 7 of Geometric Nonlinear Functional Analysis by Benyamini ...


1

The inequality does not hold in general. Let $$ T=\begin{bmatrix}1&1\\0&0\end{bmatrix},\ \ Q_1=\begin{bmatrix}0&0\\0&1\end{bmatrix}, \ \ Q_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Note that $$ ...


1

The hint is to expand the inequality you have (for which you'll have to square), and to remember that $V_n$ is a vector space. Assuming the space is real, if you expand from $\|y_{n+1}\|\leq\|y-y_{n+1}\|$ you get $$\tag{1} \langle y,y_{n+1}\rangle\leq \frac12\|y\|^2,\ \ y\in V_n. $$ Now, if we take $y=y_k$ for some $k\leq n$, $$ \langle ...


0

Suppose we're in the unit disc $\mathbb D$ for simplicity. Let $\sum_{n=0}^{\infty}a_nz^n$ be the Taylor series of $f$ in $\mathbb D.$ Using the orthogonality of the exponenetials, we see $$\int_{\mathbb D}|f|^2\, dA = \int_0^1 \int_0^{2\pi} |f(re^{it})|^2\, dt \, r\, dr = \int_0^1\int_0^{2\pi}|\sum_{n=0}^{\infty}a_nr^ne^{int}|^2\, dt\, r\, dr$$ $$ = ...


2

This is what in finite dimension is called the gradient of $f$, if it exists (and which you may call gradient in this case as well). It's the same idea as in the finite dimensional case. In order for this to work you need a natural isomorphism between the vector space and it's dual (which you usually don't have but) which is induced by the scalar product in ...


1

If $e_1,\cdots,e_k$ is an orthonormal basis then $\langle e_i,e_j\rangle$ is $1$ when $i=j$ and $0$ when $i\ne j$. Consider the case with $k=4$ and $m=3$. We have $$\lambda_1\langle e_1,e_3\rangle+\lambda_2\langle e_2,e_3\rangle+\lambda_3\langle e_3,e_3\rangle+\lambda_4\langle e_4,e_3\rangle $$ ...


2

For an orthonormal basis, $(e_n,e_m) = \delta_{nm}$. $\delta_{nm} = 1$ when $n=m$ and $0$ when $n\neq m$. What happens when you plug this into your sum?


3

The trick is to show the contrapositive. If an operator $T$ has infinite-dimensional and closed image, then it is not compact. Indeed, by restriction we get a bijective bounded linear operator $T:(\ker T)^\perp\to\text{Im}\,T$. By the open mapping theorem, $T$ maps open sets to open sets. So the image of the unit ball is open, and this is not a compact ...


0

$A^\perp = \operatorname{span}(A)^\perp = \overline{\operatorname{span}(A)}^\perp = \overline{A}^\perp$ and $A^\perp$ is closed. This is clear from the following observations: $x$ is orthogonal to $A$ iff it is orthogonal to the span of $A$, by linearity of inner products. By continuity of inner products, if $x_n \rightarrow x$ and all $x_n$ obey $\langle ...


0

Recall that the perp map is inclusion reversing. So suppose $x \in (A^{\perp})^{\perp}$ is not in $\mathrm{span}(A)$. Then $(\mathrm{span}(A))^{\perp} = A^{\perp}$ is not contained in $x^{\perp}$, so there is some element of $y \in A^{\perp}$ that is not orthogonal to $x$.


1

You are implicitly assuming that $A_0$ is injective. Let $y\in H$. Then there exists $x\in D(A_0)$ with $y=A_0x$. Then $$ \langle A_0^{-1}y,y\rangle=\langle x,A_0x\rangle\geq0. $$ Thus $A_0^{-1}$ is positive, and so there exists an orthonormal basis $\{e_n\}$ of eigenvectors. Since $A_0^{-1}$ is compact, its eigenvalues $\lambda_n$ satisfy ...


1

Yes, it is enough. Because $A_0$ is positive, symmetric and surjective, then $A_0$ is densely-defined, injective and selfadjoint. Therefore, $A_0^{-1}$ is compact, selfadjoint with trivial null space. So $A_{0}^{-1}$ has an orthnormal basis of eigenfunctions $\{e_n \}$ with corresponding eigenvalue sequence of positive numbers $$ \lambda_1 \ge ...


1

Yes. Since $Z=|PTP|=(PTPTP)^{1/2}$, it is a limit of polynomials of the form $PX_nP$, and so $PZP=Z$. The spectral projections of an operator always belong to the von Neumann algebra generated by the operator. If $\mathcal M=W^*(|PTP|)=\{|PTP|\}''$, then $E^{|PTP|}(\Delta)\in\mathcal M$ for any Borel set $\Delta$. From the first paragraph we now that ...


0

They key observation is that if $\lambda_n=0$ then $Qe_n=0$, and that $$ \ker Q=\overline{\text{span}}\,\{e_n:\ \lambda_n=0\},\ \ \ \ (\ker Q)^\perp=\overline{\text{span}}\,\{e_n:\ \lambda_n>0\}. $$ To check this, let $x\in U$. We can write, since $\{e_n\}$ is an orthonormal basis, $$ x=\sum_n\alpha_n\,e_n. $$ Then $$ ...


1

You have a linear bijection $$ Q^{1/2} : U \rightarrow U_0 $$ $Q^{1/2}$ is an isometric isomorphism because, by definition, $Q^{1/2}$ is surjective, and is injective because $\lambda_n > 0$ for all $n$, and $$ \|Q^{1/2}y\|_{U_0}=\|Q^{-1/2}Q^{1/2}y\|_{U}=\|y\|_{U}. $$ That also implies that $\{ f_n=Q^{1/2}e_n \}$ is an orthonormal ...


1

A) Have a look at https://en.wikipedia.org/wiki/Complete_metric_space ex: $\mathbb{R}$ is the completion of $\mathbb{Q}$ for the natural distance. B) Here, your scalar product defines a norm, so a distance. For this distance, the completion of $\mathcal{C}^{\infty}(S)$ is $L^2(S)$. To be more precise, any Cauchy sequence of $\mathcal{C}^{\infty}(S)$ cannot ...


1

The closed graph theorem indeed suggests itself: Suppose that $f_n\to f$ and $\phi f_n\to g$ in norm. We must show that then $g=\phi f$. This follows from $$ g(x)=\langle K_x, g\rangle = \lim\, \langle K_x, \phi f_n\rangle =\phi(x) \lim\, \langle K_x, f_n\rangle =\phi(x)f(x) . $$


1

First notice that $\overline{\mathrm{ran}(A-\lambda)}=\ker(A-\lambda)^\perp$ for $\lambda\in\mathbb{R}$. Thus, it suffices to show that $\ker(A-\lambda)=\{0\}$ and that $\mathrm{ran}(A-\lambda)$ is closed. Let $\lambda<0$. We have $$ \|(A-\lambda)u\|\|u\|\geq\langle (A-\lambda) u,u\rangle\geq -\lambda\|u\|^2 $$ for all $u\in D(A)$. Thus, ...


2

Note first that we may assume that all $T_k$ are proper isometries; because if one of them, say $T_1$, is a unitary, we get $\sum_{k=2}^NT_kT_k^*=0$, which forces $T_k=0$ for all $k\geq2$. Since $\sigma(T_k^*T_k)=\sigma(I)=\{1\}$, using that $\sigma(AB)\cup\{0\}=\sigma(BA)\cup\{0\}$ we deduce that $\sigma(T_kT_k^*)=\{0,1\}$ (the zero has to be there ...


1

Define $T$ on $\ell^2$ by $$T((x_1,x_2,\dots))=(0,x_1,x_2/2,x_3/3,\dots).$$


2

Let $\mathcal{H}=L^2[0,1]$. Let $\mathcal{D}$ be the subspace of polynomials on $[0,1]$. Let $\mathcal{D}'$ be the subspace generated by $\{ \sin(n\pi x) \}_{n=1}^{\infty}$. Both are dense, and they have nothing in common except the $0$ vector.


0

What about subspaces $\mathbb Q$ and $\mathbb R\setminus \mathbb Q$ in $\mathbb R$?


1

If $\mu$ is sigma-finite, then there are disjoint Borel subsets $\{ A_j \}_{j=1}^{\infty}$ of finite $\mu$-measure such that $\bigcup_j A_j=\Omega$. Let $f_j = T1_{A_j}$. Then $$ 1_{A_k}f_j = 1_{A_k}T1_{A_j}=T1_{A_k\cap A_j} =0 ,\;\;\; k \ne j,\\ 1_{A_j}f_j = f_j. $$ So each function $f_j$ is supported in $A_j$. Let $f$ be the a.e. unique ...


6

Let $\{a_n\}$ be a sequence of positive real numbers that increases to $1$, with the property that the sequence of products $$ a_1,\ a_1a_2,\ a_1a_2a_3,\ a_1a_2a_3a_4,\ \ldots$$ converges to a positive value. It's not hard to write down a specific example. Let $\cal H = \ell^2(\mathbf R)$ and define $T : \cal H \to \cal H$ by $$T(x_1,x_2,x_3,\ldots) = (0, ...


1

Note that $\langle Tw, Tw\rangle=\langle w, w\rangle$. Taking $w=u-v$ gives $\langle T(u-v),T(u-v)\rangle=\langle(u-v),(u-v)\rangle\implies \left| u-v\right|^2=\left|Tu-Tv\right|^2$


1

The $b_k$s must be bounded. Clearly this is a sufficient condition. It is also a necessary one: if $|b_k|$ is unbounded, then there is a subsequence $b_{k_n}$ such that $|b_{k_n}|>n$, and then we can choose $a_{k_n} = 1/b_{k_n}$ and keep the other $a_i$s zero. Then the $a_{k_n}$ are clearly square summable but $b_{k_n}a_{k_n}=1$ is not.


1

For (b)$\Rightarrow$(a), assume $\left\Vert y_{n}\right\Vert <M$. For $\epsilon>0$, choose $x_{F}\in span\left\{ x_{k}\right\} $ s.t. $\left\Vert x-x_{F}\right\Vert <\epsilon/M$. Then \begin{eqnarray*} \left|\left\langle x,y_{n}\right\rangle \right| & \leq & \left|\left\langle x-x_{F},y_{n}\right\rangle \right|+\left|\left\langle ...


3

To complement Math1000's answer, the path of working with the equality $(P+Q-PQ)^2=P+Q-PQ$ cannot lead to a proof (i.e., the argument does not work for idempotents alone). Let $$ P=\begin{bmatrix}1&1\\0&0\end{bmatrix},\ \ Q=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Then $PQ=Q$, $QP=P$, and $$P+Q-PQ=P=(P+Q-PQ)^2.$$


3

If $P+Q-PQ$ is a projection, then \begin{align} P+Q-PQ &= (P+Q-PQ)^*\\ &= P^* + Q^* - (PQ)^*\\ &= P^* + Q^* - Q^*P^*\\ &= P + Q - QP, \end{align} from which it follows that $PQ=QP$.


1

Second(*) inequality, follows from, $K=\bigcup\limits_{n\in\mathbb{N}_0}(\mathfrak{P}^{-j}+\mathfrak{p}^{-j}u(n))$. For third (**), use Parseval's identity.



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